Question

A sample of 100 students showed an average commuting time of 45 minutes and a standard deviation of 20 minutes. You want to test at 99% level of confidence whether the mean commuting time is different from 40 minutes. Compute the p-value for this test. (NOTE: USE 4 DECIMAL DIGITS)

Answer 1

To Test :-
H0 :- µ = 40
H1 :- µ ≠ 40

Test Statistic :-
t = ( X̅ - µ ) / ( S / √(n))
t = ( 45 - 40 ) / ( 20 / √(100) )
t = 2.5

Test Criteria :-
Reject null hypothesis if | t | > t(α/2, n-1)
Critical value t(α/2, n-1) = t(0.01 /2, 100-1) = 2.626 ( From t table )
| t | > t(α/2, n-1) = 2.5 < 2.626
Result :- Fail to reject null hypothesis

Decision based on P value
P - value = P ( t > 2.5 ) = 0.0141

Looking for the value t = 2.5 in t table across n - 1 = 100 - 1 = 99 degree of freedom.

Reject null hypothesis if P value < α = 0.01 level of significance
P - value = 0.0141 > 0.01, hence we fail to reject null hypothesis
Conclusion :- Fail to reject null hypothesis

There is insufficient evidence to support the claim that mean commuting time is different from 40 minutes.